-0.008 788 423 612 813 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 612 813(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.008 788 423 612 813(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.008 788 423 612 813| = 0.008 788 423 612 813


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.008 788 423 612 813.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.008 788 423 612 813 × 2 = 0 + 0.017 576 847 225 626;
  • 2) 0.017 576 847 225 626 × 2 = 0 + 0.035 153 694 451 252;
  • 3) 0.035 153 694 451 252 × 2 = 0 + 0.070 307 388 902 504;
  • 4) 0.070 307 388 902 504 × 2 = 0 + 0.140 614 777 805 008;
  • 5) 0.140 614 777 805 008 × 2 = 0 + 0.281 229 555 610 016;
  • 6) 0.281 229 555 610 016 × 2 = 0 + 0.562 459 111 220 032;
  • 7) 0.562 459 111 220 032 × 2 = 1 + 0.124 918 222 440 064;
  • 8) 0.124 918 222 440 064 × 2 = 0 + 0.249 836 444 880 128;
  • 9) 0.249 836 444 880 128 × 2 = 0 + 0.499 672 889 760 256;
  • 10) 0.499 672 889 760 256 × 2 = 0 + 0.999 345 779 520 512;
  • 11) 0.999 345 779 520 512 × 2 = 1 + 0.998 691 559 041 024;
  • 12) 0.998 691 559 041 024 × 2 = 1 + 0.997 383 118 082 048;
  • 13) 0.997 383 118 082 048 × 2 = 1 + 0.994 766 236 164 096;
  • 14) 0.994 766 236 164 096 × 2 = 1 + 0.989 532 472 328 192;
  • 15) 0.989 532 472 328 192 × 2 = 1 + 0.979 064 944 656 384;
  • 16) 0.979 064 944 656 384 × 2 = 1 + 0.958 129 889 312 768;
  • 17) 0.958 129 889 312 768 × 2 = 1 + 0.916 259 778 625 536;
  • 18) 0.916 259 778 625 536 × 2 = 1 + 0.832 519 557 251 072;
  • 19) 0.832 519 557 251 072 × 2 = 1 + 0.665 039 114 502 144;
  • 20) 0.665 039 114 502 144 × 2 = 1 + 0.330 078 229 004 288;
  • 21) 0.330 078 229 004 288 × 2 = 0 + 0.660 156 458 008 576;
  • 22) 0.660 156 458 008 576 × 2 = 1 + 0.320 312 916 017 152;
  • 23) 0.320 312 916 017 152 × 2 = 0 + 0.640 625 832 034 304;
  • 24) 0.640 625 832 034 304 × 2 = 1 + 0.281 251 664 068 608;
  • 25) 0.281 251 664 068 608 × 2 = 0 + 0.562 503 328 137 216;
  • 26) 0.562 503 328 137 216 × 2 = 1 + 0.125 006 656 274 432;
  • 27) 0.125 006 656 274 432 × 2 = 0 + 0.250 013 312 548 864;
  • 28) 0.250 013 312 548 864 × 2 = 0 + 0.500 026 625 097 728;
  • 29) 0.500 026 625 097 728 × 2 = 1 + 0.000 053 250 195 456;
  • 30) 0.000 053 250 195 456 × 2 = 0 + 0.000 106 500 390 912;
  • 31) 0.000 106 500 390 912 × 2 = 0 + 0.000 213 000 781 824;
  • 32) 0.000 213 000 781 824 × 2 = 0 + 0.000 426 001 563 648;
  • 33) 0.000 426 001 563 648 × 2 = 0 + 0.000 852 003 127 296;
  • 34) 0.000 852 003 127 296 × 2 = 0 + 0.001 704 006 254 592;
  • 35) 0.001 704 006 254 592 × 2 = 0 + 0.003 408 012 509 184;
  • 36) 0.003 408 012 509 184 × 2 = 0 + 0.006 816 025 018 368;
  • 37) 0.006 816 025 018 368 × 2 = 0 + 0.013 632 050 036 736;
  • 38) 0.013 632 050 036 736 × 2 = 0 + 0.027 264 100 073 472;
  • 39) 0.027 264 100 073 472 × 2 = 0 + 0.054 528 200 146 944;
  • 40) 0.054 528 200 146 944 × 2 = 0 + 0.109 056 400 293 888;
  • 41) 0.109 056 400 293 888 × 2 = 0 + 0.218 112 800 587 776;
  • 42) 0.218 112 800 587 776 × 2 = 0 + 0.436 225 601 175 552;
  • 43) 0.436 225 601 175 552 × 2 = 0 + 0.872 451 202 351 104;
  • 44) 0.872 451 202 351 104 × 2 = 1 + 0.744 902 404 702 208;
  • 45) 0.744 902 404 702 208 × 2 = 1 + 0.489 804 809 404 416;
  • 46) 0.489 804 809 404 416 × 2 = 0 + 0.979 609 618 808 832;
  • 47) 0.979 609 618 808 832 × 2 = 1 + 0.959 219 237 617 664;
  • 48) 0.959 219 237 617 664 × 2 = 1 + 0.918 438 475 235 328;
  • 49) 0.918 438 475 235 328 × 2 = 1 + 0.836 876 950 470 656;
  • 50) 0.836 876 950 470 656 × 2 = 1 + 0.673 753 900 941 312;
  • 51) 0.673 753 900 941 312 × 2 = 1 + 0.347 507 801 882 624;
  • 52) 0.347 507 801 882 624 × 2 = 0 + 0.695 015 603 765 248;
  • 53) 0.695 015 603 765 248 × 2 = 1 + 0.390 031 207 530 496;
  • 54) 0.390 031 207 530 496 × 2 = 0 + 0.780 062 415 060 992;
  • 55) 0.780 062 415 060 992 × 2 = 1 + 0.560 124 830 121 984;
  • 56) 0.560 124 830 121 984 × 2 = 1 + 0.120 249 660 243 968;
  • 57) 0.120 249 660 243 968 × 2 = 0 + 0.240 499 320 487 936;
  • 58) 0.240 499 320 487 936 × 2 = 0 + 0.480 998 640 975 872;
  • 59) 0.480 998 640 975 872 × 2 = 0 + 0.961 997 281 951 744;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.008 788 423 612 813(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 1011 1110 1011 000(2)

6. Positive number before normalization:

0.008 788 423 612 813(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 1011 1110 1011 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.008 788 423 612 813(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 1011 1110 1011 000(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 1011 1110 1011 000(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0000 1101 1111 0101 1000(2) × 2-7


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -7

Mantissa (not normalized):
1.0001 1111 1111 1010 1010 0100 0000 0000 0000 1101 1111 0101 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1016(10) =

011 1111 1000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 1111 1111 1010 1010 0100 0000 0000 0000 1101 1111 0101 1000 =

0001 1111 1111 1010 1010 0100 0000 0000 0000 1101 1111 0101 1000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0100 0000 0000 0000 1101 1111 0101 1000


Decimal number -0.008 788 423 612 813 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 0000 1101 1111 0101 1000

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100